What is the derivative of y=(2x+1)^3+6x(2x+1)^2 ?

The derivative of y=(2x+1)^3+6x(2x+1)^2 is 96x^2+72x+12

What is the first derivative of y=(2x+1)^3+6x(2x+1)^2 ?

The first derivative of y=(2x+1)^3+6x(2x+1)^2 is 96x^2+72x+12

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derivative of

y = (2 x + 1)^{3} + 6 x (2 x + 1)^{2}

96 x^{2} + 72 x + 12

Solution steps

\frac{d}{d x} ((2 x + 1)^{3} + 6 x (2 x + 1)^{2})

Simplify

(2 x + 1)^{3} + 6 x (2 x + 1)^{2} : 32 x^{3} + 36 x^{2} + 12 x + 1

= \frac{d}{d x} (32 x^{3} + 36 x^{2} + 12 x + 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (32 x^{3}) + \frac{d}{d x} (36 x^{2}) + \frac{d}{d x} (12 x) + \frac{d}{d x} (1)

\frac{d}{d x} (32 x^{3}) = 96 x^{2}

\frac{d}{d x} (36 x^{2}) = 72 x

\frac{d}{d x} (12 x) = 12

\frac{d}{d x} (1) = 0

= 96 x^{2} + 72 x + 12 + 0

Simplify

= 96 x^{2} + 72 x + 12

\frac{\partial}{\partial y} (e^{- 3 x} cos (6 y))

\frac{d y}{d x} = y tan (x)

\int x + 12 x^{2} d x

f (x) = (x - 2)^{2} (x - 3)^{3}

\int \frac{x + 2}{x ^{2} + x} d x

What is the derivative of y=(2x+1)^3+6x(2x+1)^2 ?
The derivative of y=(2x+1)^3+6x(2x+1)^2 is 96x^2+72x+12
What is the first derivative of y=(2x+1)^3+6x(2x+1)^2 ?
The first derivative of y=(2x+1)^3+6x(2x+1)^2 is 96x^2+72x+12